Related Rates

Water is leaking out of an inverted conical tank at a rate of 12800 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11 meters and the diameter at the top is 5 meters. If the water level is rising at a rate of 24 centimeters per minute when the height of the water is 1.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dV/dt=R−12800. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by (1/3)pi(r^2)h.

I tried to solve this, but the rate I found was not large enough to make the water level of the tank rise, so it is obviously incorrect, and I'm back where I started: confused.
Help, please???

Related rates nearly always depend on the chain rule, so you might want to try filling up this pattern...

... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (usually time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

So what have we got here? V, dV/dt, a choice of either r or h because they depend on each other, and then dh/dt the rate of increase in h. So it looks as though we should get r in terms of h rather than the other way round.

r/h = 5/22 so r = 5h/22 so the volume formula in terms of h is...

So differentiate with respect to the inner function, and the inner function with respect to t, and sub in the given values of h and dh/dt...